2.7.2-1. (a) neutral; (b) alkaline; (c) neutral; (d) acidic; (e) neutral (also, BaSO₄ is insoluble); (f) acidic; (g) neutral; (h) neutral (also, AgCl is insoluble); (i) neutral; (j) alkaline.

2.7.2-3. Weakly acidic. This is an interesting case. While HNO₃ is clearly a strong acid, Ca(OH)₂ is a medium strength base. Do ion collision analysis to prove that aqueous solutions of Ca(NO₃)₂ are slightly acidic.

2.8.3-1. (c).

2.8.3-2. True.

2.8.3-3. (b).

2.8.3-4. False. Quite often it is the H⁺ and/or OH^- ions from water or molecules of water that get discharged in an electrolysis process.

2.8.3-5. (b).

2.8.3-6. False. Electrons neither come from nowhere nor go nowhere, but are always transferred from one species to another. An oxidation process is always coupled with a reduction process and vice versa.

2.8.3-7. (c), (d), (f), and (h).

2.9.2-1. (a) S +4; O -2. (b) +1. (c) 0. (d) -1. (e) H +1; C +4; O -2. (f) H +1; N +5; O -2. (g) Na +1; S +6; O -2. (h) K +1; H +1; S +6; O -2. (i) 0. (j) K +1; Mn +7; O -2. (k) Mn +4; O -2. (l) Cu +1; I -1. (m) C +4; O -2. (n) C +2; O -2. (o) Cu +2; S +6; O -2. (p) K +1; Cr +6; O -2. (q) Cr +6; O -2; (r) Cr +6; O -2. (s) K +1; Cl +7; O -2. (t) +2. (u) S +4; O -2. (v) Fe +3; O -2. (w) Fe +2; O -2. (x) This is a tricky one! If the oxidation state of O in Fe₃O₄ is -2, then the oxidation state of Fe to make the molecule neutral should be (+2 x 4)/3 = +8/3, which is not a whole number. Can an oxidation state be a fractional number? No. However, the compound Fe₃O₄ does exist and even occurs in nature as the mineral magnetite. In fact, Fe₃O₄ is a 1:1 composition of FeO and Fe₂O₃ and is often written as FeO•Fe₂O₃. Consequently, one third of all Fe atoms in Fe₃O₄ have the oxidation state of +2 (FeO) and two-thirds have the oxidation state of +3 (Fe₂O₃). (y) Li +1; S -2. (z) H +1; O -2; Cl +1. This is one more example of a chlorine compound where the oxidation state of Cl is not -1.

2.10.3-1. (c), (e), (f), (g), and (i).

2.10.3-3. (g) disproportionation and (j) comproportionation.

2.10.3-4. Answer: False. There are only two oxidation states for calcium, 0 (Ca metal, simple substance) and +2 (any Ca compound). To disproportionate, an element should have at least three different oxidation states, so that an intermediate oxidation state can be reduced to a lower one and oxidized to a higher one.

2.10.3-5. Both types of reactions are known for these two metals. We are aware of three oxidation states for copper (0, +1, and +2) and for iron (0, +2, and +3). For each of the two metals, a compound in the intermediate oxidation state should be expected to disproportionate to compounds containing the metal in the lower and in the higher oxidation states. Likewise, compounds of Cu or Fe in the lower and in the higher oxidation state may comproportionate into a compound of the intermediate oxidation state for the metal.

2.11.5-1. (a) 0.5; (b) 0.25; (c) 0.17; (d) 3; (e) 0.58; (f) 16.6; (g) 20 (helium gas consists of monoatomic molecules); (h) 1 (there are no molecules in gold metal. Bulk metals consist of atoms or, according to the metallic bond model, of cations immersed in an electron gas, gold being no exception. We therefore consider 1 atom of Au as a molecular unit. Dividing 197 g by the atomic mass of Au (197 a.m.u.) gives the number of moles of Au.

2.11.5-2. (a) 150 g; (b) 12.6 g; (c) 200 g; (d) 243 g (see answer to 1h above); (e) 5,608 g = 5.608 kg.

2.11.5-3. (a) 1.1; (b) 0.01; (c) 50.

2.11.5-4. Xenon is a gas. Therefore, 1 mol of Xe (131.3 g) occupies the volume of 22.4 L at 0 ^oC and 1 atm. By dividing this mass in g by this volume in L we get the density in g/L: 131.1/22.4 = 5.86 g/L. As there are 1,000 mL = 1,000 cm³ in 1 L, to convert the calculated density to g/cm³ we divide the number 5.86 by 1,000, which gives 0.00586 g/cm³.

2.11.5-5. (a) 4M; (b) 4M; (c) 1M; (d) Impossible: BaSO₄ is insoluble.

2.11.5-6. First, we need to calculate the amounts of the pure reagents in the solutions. For the AgNO₃ solution, the mass percent concentration of 20% means that 100 g of this solution would contain 20 g of AgNO₃. Only 50 g of the solution was used, so we solve the proportion for x, the quantity of pure AgNO₃ in the solution used:

100 g of the solution – contains 20 g of AgNO₃
50 g of the solution – contains x g of AgNO₃

x = (50 x 20)/100 = 10 g. This is the gram amount of pure AgNO₃ used for the reaction. The number of moles in 10 g of AgNO₃ (MW = 169.9) is equal to 10/169.9 = 0.059.

100 mL of 0.5M solution of NaCl contains 0.05 mol of pure NaCl because molarity shows the number of moles of a solute in 1 L = 1,000 mL of solution.

As follows from the reaction equation AgNO₃ + NaCl = NaNO₃ + AgCl↓, the two salts react in a 1:1 molar ratio (per each mol of NaCl, 1 mol of AgNO₃ is needed). The calculated molar amount of NaCl (0.05 mol) is less than that of AgNO₃ (0.059 mol), meaning that the limiting reagent is NaCl. Consequently, the amount of AgCl produced in the reaction is 0.05 mol. Knowing the molecular weight of AgCl (143.3), we can calculate the amount of AgCl precipitated in the reaction: 143.3 x 0.05 = 7.165 g.

2.11.5-7. First, we need to calculate how much H₂SO₄ we should use. If we had to prepare 1 L of 2M H₂SO₄, we would use 2 moles of the pure acid. Since we are making one-half of that, 1 mol of H₂SO₄ should be used. The gram amount of 1 mol of H₂SO₄ is 98.1 g. You were encouraged to watch this video showing how solutions of a desired molarity are prepared in real life. Watch it. However, the case of H₂SO₄ is tricky. Normally, a solute is either directly weighed into a volumetric flask, or weighed out separately and then quantitatively transferred into a volumetric flask. Neither is an option in this particular case. Why? Because water must never be added to concentrated/pure sulfuric acid (see Volume 1). It is H₂SO₄ that must be added to water, not the other way around. I would place about 200 mL of water in a 500 mL volumetric flask, then place the flask on a scale and tare it. I would then carefully add H₂SO₄ in small portions to the flask with frequent swirling until the required amount of the acid has been added, according to the reading on the balance. Now we have about 300 g of diluted H₂SO₄ in our volumetric flask. Adding water to quite diluted H₂SO₄ is safe. So we can top off the sulfuric acid solution with water to the mark on the neck of the flask. Due to safety issues, making aqueous solutions of H₂SO₄ of an accurate molarity using concentrated sulfuric acid is more challenging than for most other substances.

2.11.5-8. Any molar concentration can be converted to mass percent concentration and vice versa, as long as the density of the solution is known. Knowing the molecular weight of a substance, we can easily convert its mole amount to grams. However, in order to convert volume to mass we need to know the density. For example, there is a 20% solution of NaOH, meaning that 100 g of this solution contains 20 g of NaOH. Since the molecular weight of NaOH is 40 a.m.u., 1 mol of NaOH is 40 g. In the given solution, there is 20 g of NaOH, which is 0.5 mol. As always, we think proportion, as outlined below.

40 g of NaOH – is 1 mol of NaOH
20 g of NaOH – is x mol of NaOH

x = (20 x 1)/40 = 0.5 mol

To calculate the molarity, we need to know the number of moles of NaOH per solution volume in liters. Do we know the number of moles per certain quantity of the solution? Yes. Do we know the volume? No. There is 0.5 mol of NaOH in 100 grams of the solution, but what is the volume of the 100 g of the solution? To answer this question, we need to know its density. The density of 20% aqueous NaOH, as can be easily found in chemistry handbooks or on the Internet, is 1.22 g/mL. Knowing both the density and the mass of our solution, we can calculate its volume: 100 g divided by 1.22 g/mL = 82 mL or 0.082 L. Now that we know both the number of moles (0.5) in the solution and its volume (0.082 L), we can calculate the molar concentration: 0.5 mol divided by 0.082 L = 6.1 mol/L.

Note that density of aqueous solutions varies in a broad range from 1 g/mL (pure water) to > 4 g/mL for Clerici solution.

2.11.5-9. (b).

2.11.5-10. (c).

2.12.6-1. (a).

2.12.6-2. (a) endothermic; (b) exothermic.

2.12.6-3. (a) because it is exothermic. In fact, KClO₃ is indeed used to make small quantities of O₂ in the laboratory.

2.12.6-4. (b). The powers to raise the concentrations to in the equation can be determined only experimentally. In fact, experiments have shown that for this reaction occurring at 100 ^oC, n = 2 and m = 0. We say that this reaction is overall second order, while being second order with respect to NO₂ and zero order with respect to CO.

2.12.6-5. In most instances, this is true. There are some reactions however, whose rates depend on the concentration of only one of two (or more) reagents involved. One such reaction is presented in problem #4 above. The rate of the reaction between NO₂ and CO is independent of the concentration of CO. There are also extremely rare cases where reaction rate is independent of all of the reagents participating in the reaction.

2.12.6-6. (a).

2.12.6-7. (c).

2.12.6-8. (b).

2.12.6-9. (b) and (d).

2.12.6-11. (b).

2.12.6-12. Equilibrium shifts to the (a) right; (b) right; (c) left; (d) right or left, depending on what factor of the two overpowers the other. Since the reaction is exothermic, raising the temperature will shift the equilibrium to the left. Raising the pressure, on the other hand, will shift the equilibrium to the right because 2 moles of NO₂ give rise to 1 mole of N₂O₄.

2.12.6-13. No, it is not. The fact that a mixture of reagents remains unchanged for a long period of time with no detectable sign of reaction does not mean that the system has reached equilibrium. It may have. But it may have not, if the reaction is too slow. This is the case with the system 2 H₂ + O₂ ⇄ 2H₂O. The platinum catalyst speeds up the forward reaction as well as the reverse reaction and, as a result, the system quickly equilibrates. The established equilibrium lies to the far right. Likewise, the lack of detectable reaction in a mixture of N₂ and H₂ at room temperature and atmospheric pressure is no indication that the reaction is at equilibrium that is shifted to the left. Likewise, the lack of reaction in a mixture of Fe and S at ambient temperature does not indicate that the mixture has reached equilibrium that lies to the left (reagents).